Step by step
The question
Question: find the directional derivative of f(x, y) = 2xy − y² at (3, 1), in the direction of ⟨1, 0⟩.
1 The partial derivatives
Term by term.
In 2xy, 2y is fixed and waits in front.
x gives 1.
So 2y.
y² has no x: it is a fixed number, so it gives 0.
Drop the zeros: ∂f/∂x = 2y.
Term by term.
In 2xy, 2x is fixed and waits in front.
y gives 1.
So 2x.
y²: the power 2 comes down: −1 times 2 is −2.
The power goes down by 1.
So 2y.
So ∂f/∂y = 2x − 2y.
2 Find the gradient
The gradient ∇f puts the partial derivatives together as one vector.
So ∇f = ⟨2y, 2x − 2y⟩.
At the point, put x = 3 and y = 1 in: ∇f(3, 1) = ⟨2, 4⟩.
3 Make it a unit vector
⟨1, 0⟩ already has length 1.
4 Dot with the gradient
The slope in direction u is the gradient dot u.
Multiply the matching parts and add.
Careful!
With ⟨1, 0⟩ itself, the answer would be 1 times too big.
2 times 1 is 2.
4 times 0 is 0.
Add: the slope is 2.
5 Check
Let's check: take a tiny step of 0.001 along u.
The change in f/0.001 is about 2.
Our answer 2 is 2.
Almost the same.
Right!
Let's look back.
First: the partial derivatives.
Then: find the gradient.
Then: make it a unit vector.
Then: dot with the gradient.
Last: check.